> However, if one enters an arithmetic expression instead of a > digit, \v behaves as if 'u' is the default scaling unit. > [...] > \v'-24000/2'
I think the behavior is correct. Any numbers that don't have a scaling factor will get the default scaling factor, and the resulting value is interpreted as units: 1 = 1v = 12000u 12p = 12000u 24000/2 = 24000v/2v = 288000000u/24000u = 12000u
